javascript - JS 测试中间件正在返回 next()

Question

我有一个像这样的中间件类

// code.js
function validation(req, res, next) {
    if (validationLogic(req)) {
        res.send(400);
        return next(false);
    }
    return next();
}

// code.test.js
describe('validation', () => {
    describe('when req is valid', () => {
        //setting up req, res, next stub

        //some other test

        //HERE IS MY QUESTION, how do I text that validation returns next(), and not next(false)
        it('return next(), and next() is called exactly once', () => {
            const spy = sinon.spy();
            nextStub = spy;
            const result = validation(reqStub, resStub, nextStub);
            assert(spy.calledOnceWithExactly());
            assert(result === nextStub()); // both of this
            assert(result === nextStub(false)); // and this line passed
        });
    });
});

我试图测试我的 validation 函数是否返回 next() 而不是 next(false)。但是在测试中，似乎只有assert(spy.calledOnceWithExactly())可以测试next中的参数。但是 assert(result === nextStub()) 后面的行除了结果实际上来自函数 next()

之外不能测试任何东西

assert(spy.calledOnceWithExactly()) 是否足够，或者是否有其他测试方法？

Answer 1

您不需要使用 spy 。您始终可以创建一个接受参数的“占位符”函数并将其替换为下一个，并确保该函数没有返回 false

it('return next(), and next() is called exactly once', () => {
    const nextStub = (arg) => (arg === false ? arg : 'I was called!');
    const result = validation(reqStub, resStub, nextStub);
    
    assert(result !== false);
    assert(result === 'I was called!');
});

类似的东西可能会起作用。或者如果你真的想使用 spy ，你可以检查 spy 的返回值和 spy 被调用的内容。确保您的 spy 函数可以接受参数。

https://sinonjs.org/releases/v7.0.0/spy-call/